Mechanical energy problem help needed!

1. The problem statement, all variables and given/known data
A car, of mass 1100 kg, is travelling down a hill, inclined at an angle of 5o to the horizontal. The driver brakes hard and skids 15 m. The coefficient of friction between the tyres and the road is 0.7. Find the initial kinetic energy and speed of the car.

Okay, here is my attempt at a solution. I would welcome any input from you guys as to whether my method is sound.
Firstly, I have to assume that the car was travelling at constant speed before the braking so that I can then calculate the tractive force of the engine:
weight of car down the hill + tractive force = friction with road
Reaction with road = 1100 x 9.8 cos 5o = 10,739 N
Weight of car down the hill = 1100 x 9.8 x sin 5o = 939.5 N
939.5 + tractive force = 0.7 x 10,739 = 7,517.3
so the tractive force of the engine before braking = 7,517.3 - 939.5 = 6577.8 N
When the car begins to skid, there is no friction between the car and the road. The only forces acting on the car are the braking friction and the weight of the car. Therefore, the work done by the brakes against the engine must be 6577.8 x 15 = 98,667 j and this must be the k.e. of the car before the brakes were applied.
1/2 x 1100 x v2 = 98,667
v = 13.4 m s-1

Sorry to revive a dead thread, but I'm currently looking at this problem and am just wondering why the weight of the car down the hill is

1100 x 9.8 x sin 5o = 939.5 N

First analyze the units. The car is 1100 kg. Acceleration due to gravity is 9.8 m/s^2. Force is in Newtons, which is kg m/s^2 (check this using F = ma). So the units are correct.

The sin 5° comes about because gravity acts straight down. You can split this F into two components -- downhill and into (normal to) the hill -- which are just F sin 5° and F cos 5°. You can remind yourself which is which by thinking of a "flat" hill, where all the force due to gravity is "into" the hill and none is "downhill." Since sin 0° = 0 and cos 0° = 1, the sin must be downhill and the cos must be into the hill.

Hence the original problem, which is the "weight" (gravity force) of the car down the hill: